int main() { char *s="nitan"; int n,i,j,c=0; char *d; n=strlen(s)/2; //printf("%d",n); for(i=1;i<=n;i++) {
if(s[n-i]!=s[n+i]) break; } d=strcpy(d,s); for(j=i;j<=n;j++) { if(d[n+j]!=d[n-j]) { c++; d[n+j]=d[n-j]; } } printf("%s %d",d,c); } On 6 September 2011 01:21, Pratz mary <pratima.m...@gmail.com> wrote: > for yahoo shudnt min additions be yahay? > > > On 6 September 2011 00:48, learner <nimish7andr...@gmail.com> wrote: > >> @sandeep Explain Algorithm/logic & Time Complexity ? >> >> Thanks >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algogeeks@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> >> > > > -- > regards Pratima :) > -- regards Pratima :) -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.